All categories

3 years ago

50 grams of water is placed into a beaker the mass of the beaker is now 115 grams what is the mass of the empty beaker

1 answer:

6 0

65 Grams is the mass of the beaker because you just take the mass of Water + Beaker and take the water away now you have the mass of the beaker

You might be interested in

In 1901, Thomas Edison invented the nickel-iron battery. The following reaction takes place in the battery. Fe(s) + 2 NiO(OH)(s)

AveGali [126]

Ans: Moles of Fe(OH)2 produced is 5.35 moles.

Given reaction:

Fe(s) + 2NiO(OH) (s) + 2H2O(l) → Fe(OH)2(s) + 2Ni(OH)2(aq)

Based on the reaction stoichiometry:

1 mole of Fe reacts with 2 moles of NiO(OH) to produce 1 mole of Fe(OH)2

It is given that there are:

5.35 moles of Fe

7.65 moles of NiO(OH)

Here the limiting reagent is Fe

Therefore, number of moles of Fe(OH)2 produced is 5.35 moles.

5 0

3 years ago

When the drug heroin comes into contact with the Marquis reagent chemical what happens?

gtnhenbr [62]

When it comes into contact with It will turn purple

5 0

3 years ago

Read 2 more answers

Chromium(III) oxide can be prepared by heating chromium(IV) oxide in vacuo at high temperature: 4Cr02 —2Cr2O3 +02 The reaction o

kkurt [141]

<u>Answer:</u> The theoretical yield and percent yield of chromium (III) oxide is 434.72 grams and 92.6 % respectively.

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

Given mass of $CrO_2$ = 480.1 g

Molar mass of $CrO_2$ = 84 g/mol

Putting values in equation 1, we get:

$\text{Moles of }CrO_2=\frac{480.1g}{84g/mol}=5.72mol$

For the given chemical equation:

$4CrO_2\rightarrow 2Cr_2O_3+O_2$

By Stoichiometry of the reaction:

4 moles of $CrO_2$ produces 2 moles of chromium (III) oxide

So, 5.72 moles of $CrO_2$ will produce = $\frac{2}{4}\times 5.72=2.86mol$ of chromium (III) oxide

Now, calculating the mass of chromium (III) oxide from equation 1, we get:

Molar mass of chromium (III) oxide = 152 g/mol

Moles of chromium (III) oxide = 2.86 moles

Putting values in equation 1, we get:

$2.86mol=\frac{\text{Mass of chromium (III) oxide}}{152g/mol}\\\\\text{Mass of chromium (III) oxide}=(2.86mol\times 152g/mol)=434.72g$

To calculate the percentage yield of chromium (III) oxide, we use the equation:

$\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100$

Experimental yield of chromium (III) oxide = 402.4 g

Theoretical yield of chromium (III) oxide = 434.72 g

Putting values in above equation, we get:

$\%\text{ yield of chromium (III) oxide}=\frac{402.4g}{434.72g}\times 100\\\\\% \text{yield of chromium (III) oxide}=\%$

Hence, the theoretical yield and percent yield of chromium (III) oxide is 434.72 grams and 92.6 % respectively.

7 0

3 years ago

How many SO2 molecules are in 1.76 mol of SO2? How many sulfur atoms and oxygen atoms are there?

Ugo [173]

Answer:

1.76 * 6.02*10^23 = 1.05952*10^24

1.05952*2 = 2.11904 *10^24 oxygen and 1.05952*10^24 sulfur atoms

5 0

2 years ago

Determine whether the bonds in CCl4 and N2 are polar or non-polar. If the bond is polar, assign the partial positive and negativ

stira [4]

Refer to the attachment

7 0

3 years ago